Question

A solution is prepared by mixing 50.0mL of 0.001 M HCl and 10.0 mL of 0.200 M NaCl. What is the molarity of chloride ions in this solution?

My work:

(0.001 mol/L)((0.05 L)= 0.005 mol HCl

(0.200 mol/L)(0.01 L)= 0.002 mol NaCl



0.005+0.002=0.007 mol of chloride ions


0.007mol/(0.05+0.01L)
=0.117 M

Is that the right method?

Answer 1

Yes, the method is right but I think you made a math error for the HCl. 0.001 x 0.05 is 5E-5